Question

Assuming an efficiency of 30.80%, calculate the actual yield of magnesium nitrate formed from 147.4g of magnesium and excess copper(II) nitrate. Write the number in grams.

Answer 1

The answer to your question is 280 g of Mg(NO₃)₂

Step-by-step explanation:

Data

Efficiency = 30.80 %

Mg(NO₃)₂ = ?

Magnesium = 147.4 g

Copper (II) nitrate = excess

Balanced Reaction

Mg + Cu(NO₃)₂ ⇒ Mg(NO₃)₂ + Cu

Reactants Elements Products

1 Mg 1

1 Cu 1

2 N 2

6 O 6

Process

1.- Calculate the theoretical yield

Molecular weight Mg = 24

Molecular weight Mg(NO₃)₂ = 24 + (14 x 2) + (16 x 6)

= 24 + 28 + 96

= 148 g

24 g of Mg -------------------- 148 g of Mg(NO₃)₂

147.4 g of Mg ------------------- x

x = (147.4 x 148) / 24

x = 908.96 g of Mg(NO₃)₂

2.- Calculate the Actual yield

yield percent =
`(actual yield)/(theoretical yield)`

Solve for actual yield

Actual yield = Yield percent x Theoretical yield

Substitution

Actual yield =
`(30.8)/(100)` x 908.96

Actual yield = 279.95 ≈ 280g